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Natali [406]
3 years ago
9

Which figure best demonstrates the setup for the box method of finding the area of a triangle

Mathematics
2 answers:
Ivan3 years ago
6 0

Answer:

The third triangle

Ivanshal [37]3 years ago
4 0
Hi there! The third option shows us the best setup.

We can find the area of a triangle with the standard formula area = 1/2 * base * height.

To be able to fill in this formula, we need to have a base and a height. We can't easily find a triangle with given base and height, so we must look for another option.

We can also take the area of a square (length × width) and then divide the answer by 2. This is possible when we take the third setup. Hence the third answer is correct.
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Step-by-step explanation:

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one vertex of a polygon is located at (3,-2). After a rotation, the vertex is located at (2,3). which transformations could have
Musya8 [376]

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One more question on this test guys. please answer as fast as possible. If you do,
GaryK [48]

1. The given rectangular equation is x=2.

We substitute x=r\cos \theta.

r\cos \theta=2

Divide through by \cos \theta

r=\frac{2}{\cos \theta}

r=2}\sec \theta

\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

We use the relation r^2=x^2+y^2 and y=r\sin \theta

This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

4. We have x=\sqrt{3}y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

\theta=\frac{\pi}{6}

\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}

5. We have x=y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta

This implies that;

\tan \theta=1

\theta=\frac{\pi}{4}

\boxed{x=y\to \theta=\frac{\pi}{4}}

6 0
3 years ago
Thea has a key on her calculator marked $\textcolor{blue}{\bf\circledast}$. If an integer is displayed, pressing the $\textcolor
mel-nik [20]

Answer:

$9$

Step-by-step explanation:

Given: Thea enters a positive integer into her calculator, then squares it, then presses the $\textcolor{blue}{\bf\circledast}$ key, then squares the result, then presses the $\textcolor{blue}{\bf\circledast}$ key again such that the calculator displays final number as $243$.

To find: number that Thea originally entered

Solution:

The final number is $243$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $243$ must be $324$.

As previously the number was squared, so the number before $324$ must be $18$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $18$ must be $81$

As previously the number was squared, so the number before $81$ must be $9$.

6 0
4 years ago
Are angles 8 and 15 <br><br> congruent <br><br> supplementary <br><br> or neither
kap26 [50]
Angles 8 and 15 are supplementary.
3 0
3 years ago
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